How do you find the equation of the tangent line to the graph of #y = (ln x)/x# at the points (1,0)?

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Answer 1 · 2016-11-04T13:08:36

The equation is #y = x - 1#.

Differentiate using the quotient rule.

#y = lnx/x#

#y' = (1/x xx x - lnx xx 1)/x^2#

#y' = (1 - lnx)/x^2#

Now, we find the slope of the tangent by inserting the point #x = 1# into the derivative.

#m_"tangent" = (1- ln(1))/1^2#

#m_"tangent" = (1 - 0)/1#

#m_"tangent" = 1#

We can now find the equation, because we know the slope and a point.

#y - y_1 = m(x - x_1)#

#y - 0 = 1(x- 1)#

#y - 0 = x - 1#

#y = x - 1#

Hopefully this helps!